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Lagrangians and integrability for additive fourth-order difference equations

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Abstract

We use a recently found method to characterise all the invertible fourth-order difference equations linear in the extremal values based on the existence of a discrete Lagrangian. We also give some result on the integrability properties of the obtained family and we put it in relation with known classifications. Finally, we discuss the continuum limits of the integrable cases.

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Notes

  1. In this section and in the next ones, we will indicate various placeholder variables with Greek letters \(\xi \), \(\eta \), \(\zeta \).... We will use these placeholders variables when making statements on functions which might have different arguments, e.g. the function \(g=g\left( \xi \right) \) in Eq. (2.5).

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Acknowledgements

GG would like to thank Prof. N. Joshi for the helpful discussions during the preparation of this paper. GG is supported by the Australian Research Council through grants FL120100094 (Prof. N. Joshi), DP190101838 (A/Prof. M. Radnović), and DP200100210 (Prof. N. Joshi and A/Prof. M. Radnović).

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Correspondence to Giorgio Gubbiotti.

Appendices

Proof of Theorem 2

Applying the method described in Sect. 2 to Eq. (2.9a), we obtain following equation (since f and h depend on the same variables we drop the explicit dependence on \(x_{n+1}\), \(x_{n}\) and \(x_{n-1}\)):

$$\begin{aligned}&{\frac{\partial ^{3} f}{\partial x_{{n+1}}^{2}\partial x_{{n-1}}}} {\frac{\partial f}{\partial x_{{n+1}}}} {\frac{\partial h}{\partial x_{{n-1}}}} - {\frac{\partial ^{3} f}{\partial x_{{n+1}}^{2}\partial x_{{n-1}}}} {\frac{\partial ^{2} h}{\partial x_{{n+1}}\partial x_{{n-1}}}}h + {\frac{\partial ^{3} h}{\partial {x_{{n+1}}}^{2}\partial x_{{n-1}}}} {\frac{\partial ^{2}f}{\partial x_{{n+1}}\partial x_{{n-1}}}} h\nonumber \\&- {\frac{\partial ^{3} h}{\partial x_{{n+1}}^{2}\partial x_{{n-1}}}} {\frac{\partial f}{\partial x_{{n+1}}}} {\frac{\partial f}{\partial x_{{n-1}}}} - {\frac{\partial ^{2} f}{\partial x_{{n+1}}^{2}}} {\frac{\partial ^{2} f}{\partial x_{{n+1}}\partial x_{{n-1}}}} {\frac{\partial h}{\partial x_{{n-1}}}}\nonumber \\&+{\frac{\partial ^{2} f}{\partial x_{{n+1}}^{2}}} {\frac{\partial ^{2} h}{\partial x_{{n+1}}\partial x_{{n-1}}}} {\frac{\partial h}{\partial x_{{n-1}}}}=0. \end{aligned}$$
(A.1)

Using the CAS Maple 2016 to solve Eq. (A.1), we find that the solution is actually independent of h and has the following form:

$$\begin{aligned} f\left( x_{n+1},x_{n},x_{n-1} \right) = G_{+}\left( x_{n+1} \right) G\left( x_{n} \right) G_{-}\left( x_{n-1} \right) . \end{aligned}$$
(A.2)

Going back to Eq. (2.9a), if we solve with respect to \(\partial ^{3} L_{n-2} /\partial x_{n-2}^{2}\partial x_{n}\) and differentiating with respect to \(x_{n+1}\), we find the following simple compatibility condition:

$$\begin{aligned}&G\left( x_{n} \right) \left[ G_{+}'(x_{n+1})\frac{\partial h}{\partial x_{n-1}} (x_{n+1}, x_{n}, x_{n-1}) -G_{+}(x_{n+1})\frac{\partial h}{\partial x_{n-1},x_{n+1}} (x_{n+1}, x_{n}, x_{n-1}) \right] \times \nonumber \\&\left[ G_{-}(x_{n-1})\frac{\partial l_{n-2}}{\partial x_{n-1},x_{n}} (x_{n}, x_{n-1}, x_{n-2}) -\frac{\partial l_{n-2}}{\partial x_{n}} (x_{n}, x_{n-1}, x_{n-2}) G_{-}'(x_{n-1}) \right] =0, \end{aligned}$$
(A.3)

where we defined:

$$\begin{aligned} l_{n-2}\left( x_{n},x_{n-1},x_{n-2} \right) \equiv \frac{\partial L_{n-2}}{\partial x_{n-2}}\left( x_{n},x_{n-1},x_{n-2} \right) . \end{aligned}$$
(A.4)

Equation (A.3) has three factors which can be annihilated separately. The first factor gives \(G\left( x_{n} \right) =0\), that is \(f\equiv 0\), which is not allowed. Therefore, from (A.3) we can choose to fix either f or \(l_{n-2}\). We will now address these two possibilities.

1.1 Fix \(l_{n-2}\) from (A.3)

Solving the second factor in (A.3), we obtain the following value for \(l_{n-2}\):

$$\begin{aligned} l_{n-2}\left( x_{n},x_{n-1},x_{n-2} \right) = l_{1,n-2}\left( x_{n},x_{n-2} \right) G_{-}\left( x_{n-1} \right) +l_{2,n-2}\left( x_{n-1},x_{n-2} \right) . \end{aligned}$$
(A.5)

Inserting (A.5) into (2.9a), we obtain the following equation:

$$\begin{aligned}&\left( G_{+} \left( x_{{n+1}} \right) G_{-}'\left( x_{{n-1}} \right) G\left( x_{{n}} \right) x_{{n-2}} +{\frac{\partial h}{\partial x_{{n-1}}}}\left( x_{{n+1}},x_{{n}},x_{{n-1}} \right) \right) \nonumber \\&\quad {\frac{\partial ^{2} l_{1,n-2}}{\partial x_{{n-2}}\partial x_{{n}}}} \left( x_{{n}},x_{{n-2}} \right) =0. \end{aligned}$$
(A.6)

Again we have two factors we can choose to annihilate. The first factors, since no function depends on \(x_{n-2}\) is equivalent to the following equations:

$$\begin{aligned} G_{+} \left( x_{{n+1}} \right) G_{-}'\left( x_{{n-1}} \right) G\left( x_{{n}} \right) = 0, \quad {\frac{\partial h}{\partial x_{{n-1}}}}\left( x_{{n+1}},x_{{n}},x_{{n-1}} \right) =0. \end{aligned}$$
(A.7)

The first equation implies \(G_{-}\left( x_{n-1} \right) =\text {constat}\) and the second one implies \(h=h\left( x_{n+1},x_{n} \right) \). This is not allowed as the equation will be independent of \(x_{n-1}\). Therefore, we are forced to annihilate the second factor. This implies:

$$\begin{aligned} l_{1,n-2}\left( x_{{n}},x_{{n-2}} \right) = l_{1,1,n-2}\left( x_{n-2} \right) +l_{1,2,n-2}\left( x_{n} \right) . \end{aligned}$$
(A.8)

Inserting this into (A.5) and using the arbitrariness of \(l_{2,n-2}\), we can write:

$$\begin{aligned} l_{n-2}\left( x_{n},x_{n-1},x_{n-2} \right) = l_{1,2,n-2}\left( x_{n} \right) G_{-}\left( x_{n-1} \right) +\frac{\partial l_{2,n-2}}{\partial x_{n-2}}\left( x_{n-1},x_{n-2} \right) . \end{aligned}$$
(A.9)

Using the definition of \(l_{n-2}\) (A.4) and the fact that discrete Lagrangians are defined only up to total difference, from formula (A.9) we obtain the following form of the Lagrangian:

$$\begin{aligned} L_{n}\left( x_{n+2},x_{n+1},x_{n} \right) = l_{1,2,n-2}\left( x_{n+2} \right) G_{-}\left( x_{n+1} \right) x_{n} +l_{2,n}\left( x_{n+1},x_{n} \right) . \end{aligned}$$
(A.10)

The Euler–Lagrange equation corresponding to (A.10), upon substitution of Eq. (2.8) are:

$$\begin{aligned}&{\frac{\partial l_{2,n}}{\partial x_{{n}}}} \left( x_{{n}},x_{{n+1}} \right) +G_{-} \left( x_{{n+1}} \right) l_{1,2,n} \left( x_{{n+2}} \right) +{\frac{\partial l_{2,n-1}}{\partial x_{{n}}}}\left( x_{{n-1}},x_{{n}} \right) \nonumber \\&\quad +G_{-}'\left( x_{{n}} \right) l_{1,2,n-1}\left( x_{{n+1}} \right) x_{{n-1}} =l_{1,2,n-2}' \left( x_{{n}} \right) \frac{ h \left( x_{{n+1}},x_{{n}},x_{{n-1}}\right) -x_{{n+2}}}{ G_{+}\left( x_{{n+1}} \right) G \left( x_{{n}} \right) }. \nonumber \\ \end{aligned}$$
(A.11)

Differentiating Eq. (A.11) with respect to \(x_{n+2}\) twice, we obtain:

$$\begin{aligned} G_{-} \left( x_{{n+1}} \right) l_{1,2,n}'' \left( x_{{n+2}} \right) =0. \end{aligned}$$
(A.12)

Using the usual argument, we obtain that we need to annihilate the second factor, which gives:

$$\begin{aligned} l_{1,2,n} \left( x_{{n+2}} \right) = C_{1,n} x_{n+2}+C_{2,n}, \end{aligned}$$
(A.13)

where \(C_{1,n}\) and \(C_{2,n}\) are two functions depending on n alone. Substituting back in Eq. (A.11) and applying the differential operator

$$\begin{aligned} \frac{\partial }{\partial x_{n}}\left( \frac{1}{G\left( x_{n} \right) } \frac{\partial }{\partial x_{n+2}} \right) , \end{aligned}$$
(A.14)

we obtain:

$$\begin{aligned} C_{1,n-2} \frac{G'\left( x_{n} \right) }{G^{2}\left( x_{n} \right) } =0. \end{aligned}$$
(A.15)

Since \(C_{1,n-2}\ne 0\), we obtain \(G\left( x_{n} \right) =1/K_{1}\) where \(K_{1}\) is a constant. Inserting this value into (A.11), we obtain:

$$\begin{aligned}&{\frac{\partial l_{2,n}}{\partial x_{{n}}}} \left( x_{{n}},x_{{n+1}} \right) +G_{-} \left( x_{{n+1}} \right) \left( C_{1,n} x_{{n+2}}+C_{2,n} \right) +{\frac{\partial l_{2,n-1}}{\partial x_{{n}}}}\left( x_{{n-1}},x_{{n}} \right) \nonumber \\&\qquad + G_{-}'\left( x_{{n}} \right) \left( C_{1,n-1}x_{{n+1}}+C_{2,n-1}\right) x_{{n-1}}\nonumber \\&\quad = C_{1,n-2} K_{1} \frac{ h \left( x_{{n+1}},x_{{n}},x_{{n-1}}\right) -x_{{n+2}}}{ G_{+}\left( x_{{n+1}} \right) }. \end{aligned}$$
(A.16)

We can take the coefficient with respect to \(x_{n+2}\), and we obtain:

$$\begin{aligned} G_{-} \left( x_{{n+1}} \right) C_{1,n} =- \frac{C_{1,n-2} K_{1}}{ G_{+}\left( x_{{n+1}} \right) }. \end{aligned}$$
(A.17)

We can rewrite this equation as:

$$\begin{aligned} G_{-} \left( x_{{n+1}} \right) G_{+}\left( x_{{n+1}} \right) =- \frac{C_{1,n-2} K_{1}}{C_{1,n}}. \end{aligned}$$
(A.18)

Since \(K_{1}\) is a constant, upon differentiation with respect to \(x_{n+1}\), there exists a constant \(q\in \mathbb {R}\setminus {\left\{ 0 \right\} }\) such that:

$$\begin{aligned} G_{+}\left( x_{{n+1}} \right) = \frac{K_{1}}{qG_{-} \left( x_{{n+1}} \right) }, \quad \text {and}\quad C_{1,n}= q C_{1,n-2}. \end{aligned}$$
(A.19)

Using conditions (A.19) into (A.16), we obtain:

$$\begin{aligned}&{\frac{\partial l_{2,n}}{\partial x_{{n}}}} \left( x_{{n}},x_{{n+1}} \right) +G_{-} \left( x_{{n+1}} \right) C_{2,n} +{\frac{\partial l_{2,n-1}}{\partial x_{{n}}}}\left( x_{{n-1}},x_{{n}} \right) \nonumber \\&\quad + G_{-}'\left( x_{{n}} \right) \left( C_{1,n-1}x_{{n+1}}+C_{2,n-1}\right) x_{{n-1}} = C_{1,n-2} G_{-}\left( x_{{n+1}} \right) h \left( x_{{n+1}},x_{{n}},x_{{n-1}}\right) . \nonumber \\ \end{aligned}$$
(A.20)

Differentiating with respect to \(x_{n+1}\) and \(x_{n-1}\), we obtain a PDE for h which can be solved to give:

$$\begin{aligned} h \left( x_{{n+1}},x_{{n}},x_{{n-1}}\right)&= \frac{h_{1}\left( x_{n},x_{n-1} \right) +h_{2}\left( x_{n+1},x_{n} \right) }{ G_{-}\left( x_{n+1} \right) }\nonumber \\&\quad -\frac{C_{1,n-1}}{qC_{1,n-2}} \frac{G_{-}\left( x_{n} \right) x_{n+1}x_{n-1}}{G_{-}\left( x_{n+1} \right) }. \end{aligned}$$
(A.21)

Since h must not depend explicitly on n, we must impose that the coefficient \(F_{n} =C_{1,n-1}/C_{1,n-2}\) is n independent, that is it is a total difference. Using again Eq. (A.19), we obtain:

$$\begin{aligned} C_{1,n-1}^{2} - q C_{1,n-2}^{2} = 0. \end{aligned}$$
(A.22)

This implies \(q>0\), that is \(q=\lambda ^{2}\) for some \(\lambda \in \mathbb {R}\setminus { \left\{ 0 \right\} }\), and then

$$\begin{aligned} C_{1,n}^{\pm } = A \left( \pm \lambda \right) ^{n} \end{aligned}$$
(A.23)

with \(A\in \mathbb {R}\) a constant. However, due to the arbitrariness of \(\lambda \) we can consider only the solution \(C_{1,n}^{+}\). Indeed, \(\lambda \) can be negative and the cases with \(C_{1,n}^{-}\) just follow from the substitution \(\lambda \rightarrow -\lambda \). Therefore, we drop the superscript \(+\) in (A.23). This reasoning implies that the h in (A.21) assumes the following form:

$$\begin{aligned} h \left( x_{{n+1}},x_{{n}},x_{{n-1}}\right) = \frac{h_{1}\left( x_{n},x_{n-1} \right) + h_{2}\left( x_{n+1},x_{n} \right) }{ G_{-}\left( x_{n+1} \right) } -\frac{G_{-}\left( x_{n} \right) x_{n+1}x_{n-1}}{\lambda G_{-}\left( x_{n+1} \right) }. \end{aligned}$$
(A.24)

We can finally insert (A.24) into (A.20) and obtain:

$$\begin{aligned}&{\frac{\partial l_{2,n}}{\partial x_{{n}}}} \left( x_{{n}},x_{{n+1}} \right) +G_{-} \left( x_{{n+1}} \right) C_{2,n} +{\frac{\partial l_{2,n-1}}{\partial x_{{n}}}}\left( x_{{n-1}},x_{{n}} \right) \nonumber \\&\quad + G_{-}'\left( x_{{n}} \right) C_{2,n-1} x_{{n-1}} = -A \lambda ^{n} \left[ f_{1}\left( x_{n},x_{n-1} \right) + f_{2}\left( x_{n+1},x_{n} \right) \right] . \end{aligned}$$
(A.25)

Differentiating with respect to \(x_{n+1}\), we obtain a linear PDE for \(l_{2,n}\left( x_{n+1},x_{n} \right) \). Solving such equation, we obtain the following form for this function:

$$\begin{aligned} l_{2,n}\left( x_{n+1},x_{n} \right)&= l_{2,2,n}\left( x_{n+1} \right) + l_{2,1}\left( x_{n} \right) - A\lambda ^{n} \int ^{x_{n}} f_{2}\left( x_{n+1}, \xi \right) {{\,\mathrm{d}\,}}\xi \nonumber \\&\quad -C_{2,n} x_{n} G_{-}\left( x_{n+1} \right) . \end{aligned}$$
(A.26)

From the form of the Lagrangian function, using the property of equivalence, we can remove the arbitrary function \(l_{2,2,n}\left( x_{n+1} \right) \) and keep only \(l_{2,1,n}\left( x_{n} \right) \). So, in (A.25) this yields:

$$\begin{aligned} \frac{l_{2,1,n}'\left( x_{n} \right) }{A \lambda ^{n}} +f_{1}\left( x_{n},x_{n-1} \right) =\frac{1}{\lambda }\int ^{x_{n-1}} \frac{\partial f_{2}}{\partial x_{n}}\left( x_{n},\xi \right) {{\,\mathrm{d}\,}}\xi . \end{aligned}$$
(A.27)

Differentiation with respect to \(x_{n-1}\) yields the following equation:

$$\begin{aligned} \frac{\partial f_{1}}{\partial x_{n-1}}\left( x_{n},x_{n-1} \right) =\frac{1}{\lambda }\frac{\partial f_{2}}{\partial x_{n}}\left( x_{n},x_{n-1} \right) . \end{aligned}$$
(A.28)

Equation (A.28) stimulates the introduction of a potential function \(V=V\left( x_{n},x_{n-1} \right) \) such that:

$$\begin{aligned} f_{1}\left( x_{n},x_{n-1} \right) = \frac{1}{\lambda } \frac{\partial V}{\partial x_{n}}\left( x_{n},x_{n-1} \right) , \quad f_{2}\left( x_{n},x_{n-1} \right) = \frac{\partial V}{\partial x_{n-1}}\left( x_{n},x_{n-1} \right) . \end{aligned}$$
(A.29)

Using such potential, we have that Eq. (A.28) is identically satisfied, while (A.27) reduces to \(l_{2,1,n}'\left( x_{n} \right) =0\). This implies \(l_{2,1,n}\left( x_{n} \right) = C_{3,n}\), but this function of n can be removed from the Lagrangian as it is a total difference.

Summing up, we obtained that if an additive fourth-order difference Eq. (1.6) is Lagrangian then it has the following form:

$$\begin{aligned}&G_{-}\left( x_{n+1} \right) x_{n+2} + \lambda ^{2} G_{-}\left( x_{n-1} \right) x_{n-2}+ \lambda G_{-}'\left( x_{n} \right) x_{n+1}x_{n-1} \nonumber \\&\quad +\frac{\partial V}{\partial x_{n}}\left( x_{n+1},x_{n} \right) +\lambda \frac{\partial V}{\partial x_{n}}\left( x_{n},x_{n-1} \right) =0. \end{aligned}$$
(A.30)

Letting \(g\equiv G_{-}\) Eq. (2.5) follows. The constant \(K_{1}\) appearing in the Lagrangian can be scaled away and we obtain the Lagrangian (2.7).

1.2 Fix h from (A.3)

If we fix h from (A.3), we obtain:

$$\begin{aligned} h\left( x_{n+1},x_{n},x_{n-1} \right) = h_{+}\left( x_{n+1},x_{n} \right) +G_{+}\left( x_{n+1} \right) h_{-}\left( x_{n},x_{n-1} \right) . \end{aligned}$$
(A.31)

After a long calculation which follows the same strategy outlined in the case when we fix \(l_{n-2}\) from (A.3) we find that this case implies \(G\equiv 0\), and so it is impossible.

This concludes the Proof of Theorem 2.

Proof of Theorem 3

Since (3.4) has to be an invariant it must satisfy the following condition:

$$\begin{aligned}&x_{n+2} P_{1}\left( x_{n+1},x_{n} \right) +x_{n-1} P_{2}\left( x_{n+1},x_{n} \right) + x_{n+2}x_{n-1} P_{3}\left( x_{n+1},x_{n} \right) + P_{4}\left( x_{n+1},x_{n} \right) \nonumber \\&\quad -x_{n+1} P_{1}\left( x_{n},x_{n-1} \right) -x_{n-2} P_{2}\left( x_{n},x_{n-1} \right) -x_{n+1}x_{n-2} P_{3}\left( x_{n},x_{n-1} \right) \nonumber \\&- P_{4}\left( x_{n},x_{n-1} \right) =0. \end{aligned}$$
(B.1)

After substituting the form of Eq. (2.5), no function depends on \(x_{n-2}\), so we can take the coefficients with respect to it. This yields the following system of functional equations which must be identically satisfied:

$$\begin{aligned}&\frac{g(x_{n-1}) P_{1}(x_{n+1}, x_{n})}{g(x_{n+1})} +\frac{g(x_{n-1}) x_{n-1} P_{3}(x_{n+1}, x_{n})}{g(x_{n+1})}\nonumber \\&\quad +P_{2}(x_{n}, x_{n-1})+x_{n+1} P_{3}(x_{n}, x_{n-1})=0, \end{aligned}$$
(B.2a)
$$\begin{aligned}&x_{n-1} P_{2}(x_{n+1}, x_{n}) -x_{n+1} P_{1}(x_{n}, x_{n-1}) +P_{4}(x_{n+1}, x_{n}) -P_{4}(x_{n}, x_{n-1}) \nonumber \\&\quad -\left( g'(x_{n}), x_{n}) x_{n-1} x_{n+1}+\frac{\partial V(x_{n+1}, x_{n})}{\partial x_{n}}+\frac{\partial V(x_{n}, x_{n-1})}{\partial x_{n}}\right) \left( \frac{P_{1}(x_{n+1}, x_{n})}{g(x_{n+1})}\right. \nonumber \\&\quad \left. + \frac{x_{n-1} P_{3}(x_{n+1}, x_{n})}{g(x_{n+1})}\right) =0. \end{aligned}$$
(B.2b)

To solve the above equations, apply again the procedure described in Sect. 2. Applying this strategy to Eq. (B.2a), we are able to solve Eq. (B.2a) fixing the form of the functions \(P_{i}\) in terms of the function g:

$$\begin{aligned} P_{1}\left( x_{n},x_{n-1} \right)&= -g(x_{n}) \left[ x_{n} (C_{1}-C_{2} x_{n-1}) g(x_{n-1})-P(x_{n-1})\right] , \end{aligned}$$
(B.3a)
$$\begin{aligned} P_{2}\left( x_{n},x_{n-1} \right)&= -\left[ x_{n-1} (C_{2} x_{n}+C_{1}) g(x_{n})+P(x_{n})\right] g(x_{n-1}), \end{aligned}$$
(B.3b)
$$\begin{aligned} P_{3}\left( x_{n},x_{n-1} \right)&= g(x_{n}) \left[ (x_{n-1}-x_{n}) C_{2}+C_{1}\right] g(x_{n-1}). \end{aligned}$$
(B.3c)

In (B.3), the function \(P=P\left( \xi \right) \) is undetermined, and \(C_{i}\) are constants. Inserting (B.3) in Eq. (B.2b), we apply the same strategy with respect to the function \(V=V\left( x_{n+1},x_{n} \right) \) and then with respect to \(P_{4}\left( x_{n+1},x_{n} \right) \). After some steps, we find the following equation:

$$\begin{aligned} C_{2} g\left( x_{n} \right) g'\left( x_{n} \right) =0. \end{aligned}$$
(B.4)

This equation implies \(C_{2}=0\), as otherwise the function \(g=g\left( \xi \right) \) will be a trivial constant. Substituting \(C_{2}=0\), we obtain the following PDE for \(V=V\left( x_{n},x_{n-1} \right) \):

$$\begin{aligned} -C_{1} g''(x_{n-1}) x_{n} + 2 C_{1} g'(x_{n}) -C_{1} \frac{\partial V(x_{n}, x_{n-1})}{\partial *{2}{x_{n-1}}, x_{n}} +P''(x_{n-1}) =0. \end{aligned}$$
(B.5)

giving the form of V in terms of g and P:

$$\begin{aligned} V(x_{n}, x_{n-1})&= V_{3}(x_{n})+V_{2}(x_{n}) x_{n-1}+V_{1}(x_{n-1}) -\frac{x_{n}^2}{2} g(x_{n-1}) +g(x_{n}) x_{n-1}^2\nonumber \\&\quad +\frac{x_{n}}{C_{1}} P(x_{n-1}). \end{aligned}$$
(B.6)

By arbitrariness of \(V_{1}\left( x_{n-1} \right) \), we can write \(V_{1}\left( x_{n-1} \right) =W\left( x_{n-1} \right) - V_{3}\left( x_{n-1} \right) \) and remove the total difference \(V_{3}\left( x_{n} \right) -V_{3}\left( x_{n-1} \right) \). That is, we can write \(V\left( x_{n},x_{n-1} \right) \) as:

$$\begin{aligned} V(x_{n}, x_{n-1})= V_{2}(x_{n}) x_{n-1}+W(x_{n-1}) -\frac{x_{n}^2}{2} g(x_{n-1}) +g(x_{n}) x_{n-1}^2+\frac{x_{n}}{C_{1}} P(x_{n-1}). \end{aligned}$$
(B.7)

Going back to Eq. (B.2b) and removing iteratively all the functions depending on \(x_{n+1}\) and \(x_{n-1}\), we finally find the following condition on g:

$$\begin{aligned} 3C_1 g'''(x_{n}) g(x_{n})^3 = 0. \end{aligned}$$
(B.8)

As g needs to be non-trivial and \(C_{1}\ne 0\) from (B.7) we finally obtain that g has to be second-order polynomial of the form (3.5).

Using the conditions in (B.2b), we find the following expression for the function \(V_{2}\):

$$\begin{aligned} V_{2}\left( x_{n} \right) = \frac{A_{4}}{\sqrt{A_{1} x_{n}^2+A_{2} x_{n}+A_{3}}} +\frac{3 A_{2} C_{1} x_{n}^2+2 C_{5} x_{n}+2 C_{6}}{2C_{1}}. \end{aligned}$$
(B.9)

The function \(V_{2}\left( x_{n} \right) \) appears to be algebraic in \(x_{n}\). However, substituting back in order to check the compatibility conditions we obtain \(A_{4}=0\). Therefore, no algebraic term is left.

The above computations produce rather cumbersome expression for \(P_{4}\left( x_{n},x_{n-1} \right) \), which we will not present. However, we notice that this final form of \(P_{4}\left( x_{n},x_{n-1} \right) \) yield the following condition for the function \(W=W\left( \eta \right) \):

$$\begin{aligned} W'\left( \eta \right) = \frac{1}{C_{1}} \frac{C_{1} (A_{2}^2 -A_{1} A_{3}) \eta ^3-(A_{1} C_{6}-A_{2} C_{5}) \eta ^2+A_{6} C_{1} \eta +A_{5} C_{1}}{A_{1} \eta ^2+A_{2} \eta +A_{3}} \end{aligned}$$
(B.10)

Since \(C_{1}\ne 0\) we perform the scaling \(C_{5}=A_{7}C_{1}\) and \(C_{6}=A_{8}C_{1}\). This yields the expression (3.7) for \(W\left( \eta \right) \) and shows that

$$\begin{aligned} P_{4}\left( x_{n},x_{n-1} \right)= & {} -x_{n-1}^2 g\left( x_{n} \right) \left[ (A_1 x_{n}+A_2) x_{n-1}+ (2 A_2 x_{n}+A_7) \right] \nonumber \\&-\left[ (A_1 A_3+A_2^2) x_{n}^3+A_2 (2 A_3+A_7) x_{n}^2\right. \nonumber \\&\left. +(A_2 A_8+2 A_3^2+A_6) x_{n}+A_3 A_8+A_5\right] x_{n-1} \nonumber \\&-x_{n} (A_2 A_3 x_{n}^2+A_3 A_7 x_{n}+A_3 A_8+A_5). \end{aligned}$$
(B.11)

This concludes the Proof of Theorem 3.

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Gubbiotti, G. Lagrangians and integrability for additive fourth-order difference equations. Eur. Phys. J. Plus 135, 853 (2020). https://doi.org/10.1140/epjp/s13360-020-00858-y

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